Brightness of a Real Image

On this page we treat the case of a camera viewing a real image
produced by a positive lens, with the scene on the other side of the
lens from the camera. We show that the brightness (per unit area)
of the image formed in the camera is unchanged by the lens.

On this page, we will be assuming that the image brightness in a camera, when viewing a scene directly, is independent of the distance from the camera to the scene. That was shown when we discussed camera image brightness.

We'll
now present a geometric argument that the image brightness is not
affected by a simple positive lens producing a real image. The
argument, though purely intuitive, is somewhat labored, which
detracts substantially from its appeal (which is sad, considering
how long it took to do the illustrations for it). Consequently,
we'll also provide a simple algebraic proof down at the end of the page.

In figure 1 we show the camera, lens, and scene (which is a magenta arrow). The lens, labeled "Lens 1" (vertical blue line near the right side) focuses the light from the image, forming a "real image", which we'll call the lens image. That is in turn focused by the camera's lens to form the image in the camera, the "camera image". The distance from the scene to the lens is l1, the distance from Lens 1 to the "lens image" is l2, and the distance from the lens image to the camera is l3. The magenta lines show the path of the light which enters the camera from an infinitesimal part of the scene, δv, located at the base of the arrow.

We've traced the path of the light from the region δv from the scene, through Lens 1, and on to the camera.

Cone 3 is the cone of light actually entering the camera from the projected "lens image" of δv. (The lenses are, of course, disks -- they're not really just line segments, as they're drawn! Hence the bundle of light from δv forms a cone, not a triangle.)

Cone 2
is the path the light followed going from Lens 1 to the lens image.
Cone 2 was drawn by simply extending the lines from cone 3 all
the way back to Lens 1.

Cone 1 is the path which was followed by that light as it went from δv in the scene to Lens 1. It was formed by observing that all the light entering cone2 must have come from δv, so we can just draw lines from the base of cone 2 to the scene. Light from outside cone 1, conversely, won't enter the camera, as it won't enter cone 2, and hence can't go through the projected image of δv and into cone 3 on a path leading to the camera's lens.

Figure 1: Scene, Lens image, Camera image equidistant

In figure 1, we've arranged the scene, lens, and camera such that the scene is the same distance from the
lens as the projected image, and the camera and lens are the same
distance from the projected image. So, we have:

(1)

By
construction, the apex angle of Cone 2 is equal to the apex angle of
Cone 1 (which is φ). Since Cone 2 and Cone 3 are the same
height, and have the same angle, they are identical. Since Cone 1
and Cone 2 have the same base size, and they are also the same height,
they're also identical. So, Cone 1, Cone 2, and Cone 3 are all
identical in figure 1.

Consequently, in figure 1, we also have

(2)

The amount of light passing through Lens 1 and going on to the camera is the light from δv
which enters Cone 1. Since Cone 1 is identical to Cone 3, that's
identical to the amount of light that would enter the camera if the
scene itself were at the location of the lens image. Since the
scene and the lens image are the same size in figure 1, we conclude
that the camera image brightness is the same as it would be if the
camera were "viewing" the scene directly.

I've used a lot of
words here, which may have obscured the basic simplicity of the
situation in figure 1. We've contrived, by keeping l1, l2, and l3
the same length, to make it completely clear that the lens is making no
difference to the brightness in this case. If this doesn't seem
clear, just forget the text and look at figure 1, and think about how
the light goes.

In figure 2, we've allowed the distance from the lens image to the camera to vary. (It's shown as smaller than l1andl2 in the figure, but it could just as well have been larger; the point is it's no longer fixed.)

As long as we keep l1andl2
equal, the size of the lens image and the scene will be the same.
Furthermore, since Cone 1 and Cone 2 are identical in this case, and they both must have the same apex angle as cone 3, the light passing through the lens image of δv and going on to the camera doesn't depend on the value of l1andl2. So, since the amount of light received from δv doesn't depend on l1andl2,
and the size of the image doesn't depend on them, the image brightness
can't depend on them either. So, the brightness will be unchanged
if we set l1=l2 = l3, and that brings us back to the case shown in figure 1.

Finally, in figure 3 we allow l1
to vary as well. (Relative placement of Lens 1 between the scene
and camera is now unconstrained.) Looking at the figure, it's
simplest to think of what we're doing here as dragging the scene to the
right or the left, while varying the height of the scene in
order to keep the lens image and camera images unchanged. Does
the brightness of the camera image change as we do that?

The fraction of the light gathered into the camera from δv varies inversely with the square ofl1. But the magnification also varies inversely with l1. Specifically, the magnification of the lens image is l2/l1, so the area of the image of δv will vary as the square of l2/l1.
Since the brightness varies as the ratio of the total light
gathered to the area over which it's spread, the net brightness won't
be affected.

Figure 3: Varying length l3and length l1

A Symbolic Proof

That
completes the "geometric" proof. If you got lost in the English
explanations accompanying the figures, or found my explanations to be
hopelessly unclear, you may have found the proof less than
satisfactory. So we'll now redo it purely algebraically.

The total light emitted from δv will be δE. We'll assume that the light is radiated evenly over a full hemisphere. This is the same assumption we made on the camera image brightness
page. With that assumption, and with Lens 1 in place, we're going
to compute the brightness of the camera image and see if it's the same
as the image brightness for the camera alone.

Light gathered by the camera will be the solid angle
subtended by Cone 1 divided by the solid angle of a full hemisphere,
times the light emitted by the image. We'll call that amount δEe. We're going to assume that the scene is distant relative to the lens diameter, so we can use small-angle approximations. Then we have:

(3)

The area of the camera image of δv will be the square of the magnification due to Lens 1 times the magnification due to the camera's lens, or

(4)

From our page on ideal lenses, case (1), we can write the magnifications as

(5)

Looking at figure 3, we can see that we can write p1 as

(6)

Now we divide (3) by (4), and plug in the values from (5) and (6), to obtain,

(7)

Almost everything cancels; after taking the limit in small δv, we're left with:

(8)

which is the same formula we found in the simple camera brightness page, equation (SC-6), which was to be shown.